le Châtelier's Principle: Precipitation Reactions

The change in meaning from soluble to insoluble is indistinctA precipitation reaction is one that results in the formation of an insoluble product when solutions of ions are mixed.  The term insoluble is one that needs careful definition.  It is the opposite of the word soluble, yet neither word has a very specific meaning.  The terms soluble and insoluble are qualitative, not quantitative.  In one sense, everything is soluble, since every substance will dissolve in another substance a little bit.   For example, we don't normally think that glass dissolves in water.  Each time you take a drink from a glass, however, a few "molecules" of glass do indeed dissolve into the water.  However, the number of molecules is so small that you don't really need to worry about it.

So, a semi-quantitative definition of the terms might be useful.  On that basis, let's agree that if a substance dissolves more than 0.1 mol/L, (0.1 molar) that it is soluble.  If it dissolves less than that it is insoluble.  Notice though that in the picture on the right there is no sharp dividing line between soluble and insoluble substances.

When we react aqueous ionic solutions the products may form a solid that settles from the solution.  This will happen only if the solubility of the product is low enough – in other words if the product is insoluble enough – so that we create a saturated solution.  Even a quite insoluble substance may not precipitate if the concentration of ions is low. 

We could write a general equation for the solution equilibrium of any ionic substance in water as:

AX (s) A+ (aq) + X- (aq)

We know that the concentration of a solid is fixed, so we can't change this equilibrium by adding more AX (s).  However, we can use le Châtelier's principle to predict what should happen as we change the concentration of the ions in solution.

If the concentration of the A+ (aq) ions, or the  X- (aq) ions, or both, were increased then we would expect (from le Châtelier's principle) to see:
less precipitate
more precipitate
no change in the precipitate

But how can you change the concentration of either the A+ (aq) ions, or the  X- (aq) ions?  You can't do it by adding more AX (s).  This is a saturated solution, and so no more AX (s) can dissolve.

There are two possibilities:

If water was evaporated, then we would expect (from le Châtelier's principle):
less solid AX
more solid AX
no change in the amount of AX

Adding the same ion from a different source is called a "common ion".   This is a standard chemical procedure, and the change predicted by le Châtelier's principle when we mix the same ions from different sources, is called the "common ion effect".  For example, for our imaginary compound AX (s) we would have a common ion of X- if we dissolved some NaX in the solution.  We would have a common ion of A+ if we dissolved some ACl in the solution.

Consider this solution equation.

AgCl (s) Ag+ (aq) + Cl- (aq)

Which of the following solids when dissolved would cause the common ion effect in a saturated solution of AgCl (s)?
Ag2SO4    NaCl K2SO4
MgCl2 all of them none of them

In this experiment you will examine the common ion effect on a simple solution equilibrium.

Materials:

Procedures:

The equilibrium you will be investigating is the solubility of Ca(OH)2 (s) in water.  Ca(OH)2 (s) is used because it is right on the borderline between our definition of soluble and insoluble.

Caution:

Sodium hydroxide is an extremely corrosive substance.  Solid NaOH, and its solutions must not come in contact with your skin.  Contact with your eyes could cause blindness.  If you get any on your skin or in your eyes, wash for at least 15 minutes with running water.

Calcium hydroxide, also known as "slaked lime", is a fairly safe chemical, but it is somewhat corrosive.  It could cause damage to your eyes, and will be midly corrosive to your skin.  You need to observe the normal safety precautions you would for any lab while using it.

Wear goggles while doing this experiment.

1. Get about 10 mL of a saturated solution of Ca(OH)2 in a clean, dry 18 x 150 mm test tube.  Your teacher will have prepared this solution ahead of time for you, to allow it to become saturated.
2. Filter the Ca(OH)2 solution into another clean, dry test tube.  This will ensure that it contains no Ca(OH)2 (s).  The resulting solution should be clear.  This solution, though it will contain no solid, is still a saturated solution since it will not be able to dissolve any more Ca(OH)2 (s).
3. Observe the effect when you add one pellet of solid NaOH.  Stopper the tube with a rubber stopper, shake, and observe the contents.

Conclusion:

1. Write out the solubility equation for Ca(OH)2 (s).  Which of the following is correct?

a) Ca(OH)2 (s) Ca (aq)  + 2OH (aq)
b) Ca(OH)2 (s) Ca+ (aq)  + (OH)-2 (aq)
c) Ca(OH)2 (s) Ca+ (aq)  + 2OH- (aq)
d) Ca(OH)2 (s) Ca2+ (aq)  + 2OH- (aq)

2. In this experiment, we make the assumption that NaOH (s) will dissolve in the solution just as it would in water.  NaOH (s) is highly soluble in water, so it should dissolve well.   Write out the solubility equation for NaOH (s) and then answer this question:  What is the common ion?

a) hydroxide b) sodium c) calcium d) chloride
Demonstration of the common ion effect on solubility

The colorless solution on the left is pure Ca(OH)2.  The one on the right has one added pellet of NaOH.

3. Which of the following statements is the best explanation for the results observed in this experiment?

a) The solid NaOH cannot dissolve very well, since it has a common ion with the Ca(OH)2.   This forces the solubility of the NaOH to the left so that it forms a cloudy, rather than a clear solution.
b) The solid NaOH dissolves in water to produce Na+ ions.  These ions reduce the solubility of the Ca(OH)2, since there are less available water molecules to hydrate the calcium hydroxide.  Therefore the calcium hydroxide solubility is reduced.
c) The solid NaOH dissolves in water to produce OH- ions.  A higher concentration of hydroxide ions forces the reaction to the left, producing more solid Ca(OH)2 as a precipitate.